Although I have used the UK National Lottery as an example my question really goes much further.
The lottery is roughly 14m - 1 per ticket. It is a 6 from 49 random draw. You are allowed to pick 6 numbers per ticket and must match them all (in no particular order) to win the jackpot.
My question is:
If you buy 7 random tickets (all different combinations) then odds seem to be 7/14m = just under 2m.
If you picked 7 numbers and then permed all '6 group' combinations (7) odds = just under 2m.
This is the part I dont really get...:
Would it be better to have 7 tickets entered in a 14m - 1 shot, OR 1 ticket entered in a 2m - 1 shot.
I contend that although it appears that they are the same odds the 1 ticket in 2m is the better bet.
Because the 7 tickets in the '14m - 1' shot all have a '14m - 1' chance of winning. So you have 7 chances at '14m to 1'. Or to put it another way there are 13,999, 993 chances of losing.
Whereas the 1 ticket in a 2m to 1 shot has a 2m to 1 chance of winning or 1,999,999 chances of losing.
Therefore the 1 ticket in the 2m lottery has a far lower number of non winning chances... (around 12,000,000 less chances!) so it must be a better bet...
Am I right???
any math proffessors out there or where could I look. I will admit that the math would seem to say that they are even, but if they are I cant understand how.